Linear First Order ODE/y' = x + y

Theorem

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = x + y$

has the general solution:

$y = C e^x - x - 1$

Initial Value: $\map y 0 = 1$

The linear first order ODE:

$(1): \quad \dfrac {\d y} {\d x} = x + y$

with initial condition:

$\map y 0 = 1$

has the particular solution:

$y = 2 e^x - x - 1$

Proof

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - y = x$

$(2)$ is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x = -1$

$\map Q x = x$

Thus:

\(\ds \int \map P x \rd x\)

\(=\)

\(\ds -\int \rd x\)

\(\ds \)

\(=\)

\(\ds - x\)

\(\ds \leadsto \ \ \)

\(\ds e^{\int P \rd x}\)

\(=\)

\(\ds e^{-x}\)

\(\ds \)

\(=\)

\(\ds \frac 1 {e^x}\)

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

\(\ds \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }\)

\(=\)

\(\ds \frac x {e^x}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac y {e^x}\)

\(=\)

\(\ds \int \frac x {e^x} \rd x\)

\(\ds \)

\(=\)

\(\ds -\frac 1 {e^x} \paren {x + 1} + C\)

Primitive of $x e^{a x}$

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C e^x - x - 1\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Appendix $\text{A}$. Numerical Methods